The overall average salary for all employees in an office is ₹16,000. Officers earn an average salary of ₹28,000, while non-officers make an average of ₹10,000. If there are 15 officers, determine the number of non-officers.

Correct — A. Balance what the officers are above the overall average against what the non-officers are below it.

Each officer above ₹16,000: 28,000 − 16,000 = ₹12,000
Total excess of 15 officers: 15 × 12,000 = ₹1,80,000

Each non-officer below ₹16,000: 16,000 − 10,000 = ₹6,000
Non-officers needed to absorb the excess: 1,80,000 ÷ 6,000 = 30 → option (a)

Why the others are wrong

• (b)
  40 — With 40 non-officers the average falls to about ₹14,909: (4,20,000 + 4,00,000) ÷ 55. That is below ₹16,000, so 40 is too many.
• (c)
  50 — 50 non-officers give (4,20,000 + 5,00,000) ÷ 65 ≈ ₹14,154, well under the ₹16,000 average. More low earners pull the average further down.
• (d)
  60 — 60 non-officers give (4,20,000 + 6,00,000) ÷ 75 = ₹13,600. That is a 1 : 4 split of officers to non-officers, when the salary gaps call for 1 : 2.

At the overall average, what one group is above it equals what the other is below it. Each officer is ₹12,000 above ₹16,000 and each non-officer ₹6,000 below.

So headcounts go in the inverse ratio of those gaps: officers : non-officers = 6,000 : 12,000 = 1 : 2. With 15 officers, there are 30 non-officers.

Key facts

• Alligation: n₁ : n₂ = (A − a₂) : (a₁ − A), where a₁ > A > a₂, a₁ and a₂ are the group averages and A is the combined one.
• The combined average always lies between the two group averages.
• The group whose average is farther from the combined average is the smaller group.

Common traps

• Inverting the gaps and writing officers : non-officers = 2 : 1, which gives 7.5 non-officers.
• Averaging ₹28,000 and ₹10,000 to ₹19,000, as if the two groups were the same size.

No directly related past PYQ was found.